Chapter_23_Solutions - 23.1: 1 1 1 1 = 0.357 J U = kq1q2 -...

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23.1: U = kq 1 q 2 1 r 2 1 r 1 ⎟ = k(2.40 µ C)( 4.30 C) 1 0.354m 1 0.150m ⎟ = 0.357 J W =−∆ U =− 0.357 J.
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23.2: W =− 1.9 × 10 8 J =−∆ U = U i U f U f = 1.9 × 10 8 J + 5.4 × 10 8 J = 7.3 × 10 8 J
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23.3: a) E i = K i + U i = 1 2 (0.0015 kg)(22.0 m s ) 2 + k(2.80 × 10 6 C)(7.50 × 10 6 C) 0.800 m = 0.608 J E i = E f = 1 2 mv f 2 + kq 1 q 2 r f v f = 2(0.608 J 0.491 J) 0.0015 kg = 12.5 m s . b) At the closest point, the velocity is zero: 0.608 J = kq 1 q 2 r r = k(2.80 × 10 6 C)(7.80 × 10 6 0.608 J = 0.323 m.
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23.4: U =− 0.400 J = kq 1 q 2 r r = k(2.30 × 10 6 C)(7.20 × 10 6 C) 0.400 J = 0.373 m.
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23.5: a) U = kQq r = k(4.60 × 10 6 C) (1.20 × 10 6 C) 0.250 m = 0.199 J. b) (i) K f = K i + U i U f = 0J + k(4.60 × 10 6 C) (1.20 × 10 6 C) 1 0.25 m 1 0.5 m = 0.0994 J K f = 0.0994 J = 1 2 mv f 2 v f = 2(0.0994 J) 2.80 × 10 4 kg = 26.6 m s. (ii) K f = 0.189 J, v f = 36.7 m s. (iii) K f = 0.198 J, v f = 37.6 m s.
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23.6: U = kq 2 0.500m + 2kq 2 0.500 m = 6kq 2 = 6k(1.2 × 10 6 C) 2 = 0.078 J.
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23.7: a) U = k q 1 q 2 r 12 + q 1 q 3 r 13 + q 2 q 3 r 23 = k (4.00 nC)( 3.00 nC) (0.200 m) + (4.00 nC)(2.00 nC) (0.100 m) + ( 3.00 nC)(2.00 nC) (0.100 m) ⎟ =− 3.60 × 10 7 J. b) If U = 0, 0 = k q 1 q 2 r 12 + q 1 q 3 x + q 2 q 3 r 12 x . So solving for x we find: 0 =− 60 + 8 x 6 0.2 x 60x 2 26x + 1.6 = 0 x = 0.074 m, 0.360 m. Therefore x = 0.074 m since it is the only value between the two charges.
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23.8: From Example 23.1, the initial energy E can be calculated: i E i = K i + U i = 1 2 (9.11 × 10 31 kg)(3.00 × 10 6 ms ) 2 + k( 1.60 × 10 19 C)(3.20 × 10 19 C) 10 10 m E i =− 5.09 × 10 19 J. When velocity equals zero, all energy is electric potential energy, so: 5.09 × 10 19 J k2e 2 r r = 9.06 × 10 10 m.
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23.9: Since the work done is zero, the sum of the work to bring in the two equal charges q must equal the work done in bringing in charge Q . W qq = W qQ ⇒− kq 2 d = 2kqQ d Q =− q 2 .
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23.10: The work is the potential energy of the combination. U = U p + U pe + U e = ke(2e) 52 × 10 10 m + ke( e) 5 × 10 10 m + k( e)(2e) 5 × 10 10 m = ke 2 5 × 10 10 m 2 2 1 2 = (9.0 × 10 9 Nm 2 C 2 )(1.6 × 10 19 C) 2 5 × 10 10 m 2 2 3 =− 7.31 × 10 19 J Since U is negative, we must do + 7.31 × 10 19 J to separate the particles
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23.11: K 1 + U 1 = K 2 + U 2 ; K 1 = U 2 = 0 so K 2 = U 1 U 1 = e 2 4 π 0 1 r + 2 r + 2 r = 1 4 πε 0 5e 2 r , with r = 8.00 × 10 10 m U 1 = 1.44 × 10 18 J = 9.00 eV
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